1. Introduction; Selection, Projection. 2. Cartesian Product, Join. 5. Formal Definitions, A Bit of Theory

Transcription

1 Relational Algebra 169 After completing this chapter, you should be able to enumerate and explain the operations of relational algebra (there is a core of 5 relational algebra operators), write relational algebra queries of the type join select project, discuss correctness and equivalence of given relational algebra queries. Relational Algebra 170 Overview 1. Introduction; Selection, Projection 2. Cartesian Product, Join 3. Set Operations 4. Outer Join 5. Formal Definitions, A Bit of Theory

2 Example Database (recap) 171 STUDENTS SID FIRST LAST 101 Ann Smith Michael Jones (null) 103 Richard Turner Maria Brown... EXERCISES CAT ENO TOPIC MAXPT H 1 Rel.Alg. 10 H 2 SQL 10 M 1 SQL 14 RESULTS SID CAT ENO POINTS 101 H H M H H M H M 1 7 Relational Algebra (1) 172 Relational algebra (RA) is a query language for the relational model with a solid theoretical foundation. Relational algebra is not visible at the user interface level (not in any commercial RDBMS, at least). However, almost any RDBMS uses RA to represent queries internally (for query optimization and execution). Knowledge of relational algebra will help in understanding SQL and relational database systems in general.

3 Relational Algebra (2) 173 In mathematics, an algebra is a set (the carrier), and operations that are closed with respect to the set. Example: (N, {, +}) forms an algebra. In case of RA, the carrier is the set of all finite relations. We will get to know the operations of RA in the sequel (one such operation is, for example, ). Relational Algebra (3) 174 Another operation of relational algebra is selection. In contrast to operations like + in N, the selection σ is parameterized by a simple predicate. For example, the operation σ SID=101 selects all tuples in the input relation which have the value 101 in column SID. Relational algebra: selection σ SID=101 RESULTS SID CAT ENO POINTS 101 H H M H H M H M 1 7 = SID CAT ENO POINTS 101 H H M 1 12

4 Relational Algebra (4) 175 Since the output of any RA operation is some relation R again, R may be the input for another RA operation. The operations of RA nest to arbitrary depth such that complex queries can be evaluated. The final results will always be a relation. A query is a term (or expression) in this relational algebra. A query π FIRST,LAST (STUDENTS σ CAT= M (RESULTS)) Relational Algebra (5) 176 There are some difference between the two query languages RA and SQL: Null values are usually excluded in the definition of relational algebra, except when operations like outer join are defined. Relational algebra treats relations as sets, i.e., duplicate tuples will never occur in the input/output relations of an RA operator. Remember: In SQL, relations are multisets (bags) and may contain duplicates. Duplicate elimination is explicit in SQL (SELECT DISTINCT).

5 Relational Algebra (6) 177 Relational algebra is the query language when it comes to the study of relational query languages (DB Theory): The semantics of RA is much simpler than that of SQL. RA features five basic operations (and can be completely defined on a single page, if you will). RA is also a yardstick for measuring the expressiveness of query languages. If a query language QL can express all possible RA queries, then QL is said to be relationally complete. SQL is relationally complete. Vice versa, every SQL query (without null values, aggregation, and duplicates) can also be written in RA. Selection (1) 178 Selection The selection σ ϕ selects a subset of the tuples of a relation, namely those which satisfy predicate ϕ. Selections acts like a filter on a set. Selection σ A=1 A B A B =

6 Selection (2) 179 A simple selection predicate ϕ has the form Term ComparisonOperator Term. Term is an expression that can be evaluated to a data value for a given tuple: an attribute name, a constant value, an expression built from attributes, constants, and data type operations like +,,, /. Selection (3) 180 ComparisonOperator is = (equals), (not equals), < (less than), > (greater than),,, or other data type-dependent predicates (e.g., LIKE). Examples for simple selection predicates: LAST = Smith POINTS 8 POINTS = MAXPT.

7 Selection (4) 181 σ ϕ (R) may be imlemented as: Naive selection create a new temporary relation T ; foreach t R do p ϕ(t); if p then insert t into T ; fi od return T ; If index structures are present (e.g., a B-tree index), it is possible to evaluate σ ϕ (R) without reading every tuple of R. Selection (5) 182 A few corner cases σ C=1 σ A=A σ 1=2 A B A B A B = = (schema error) A B = A B

8 Selection (6) 183 σ ϕ (R) corresponds to the following SQL query: SELECT * FROM R WHERE ϕ A different relational algebra operation called projection corresponds to the SELECT clause. Source of confusion. Selection (7) 184 More complex selection predicates may be performed using the Boolean connectives: ϕ 1 ϕ 2 ( and ), ϕ 1 ϕ 2 ( or ), ϕ 1 ( not ). Note: σ ϕ1 ϕ 2 (R) = σ ϕ1 (σ ϕ2 (R)). and Are the Boolean connectives, strictly needed? The selection predicate must permit evaluation for each input tuple in isolation. Thus, exists ( ) and for all ( ) or nested relational algebra queries are not permitted in selection predicates. Actually, such predicates do not add to the expressiveness of RA.

9 Projection (1) 185 Projection The projection π L eliminates all attributes (columns) of the input relation but those mentioned in the projection list L. Projection π A,C A B C = A C Projection (2) 186 The projection π Ai1,...,A ik (R) produces for each input tuple (A 1 : d 1,..., A n : d n ) an output tuple (A i1 : d i1,..., A ik : d ik ). π may be used to reorder columns. σ discards rows, π discards columns. DB slang: All attributes not in L are projected away.

10 Projection (3) 187 In general, the cardinalities of the input and output relations are not equal. Projection eliminates duplicates A B π B = B 4 5 Projection (4) 188 π Ai1,...,A ik (R) may be imlemented as: Naive projection create a new temporary relation T ; foreach t = (A 1 : d 1,..., A n : d n ) R do u (A i1 : d i1,..., A ik : d ik ); insert u into T ; od eliminate duplicate tuples in T ; return T ; The necessary duplicate elimination makes π L one of the more costly operations in RDBMSs. Thus, query optimizers try hard to prove that the duplicate eliminaton step is not necessary.

11 Projection (5) 189 If RA is used to formalize the semantics of SQL, the format of the projection list is often generalized: Attribute renaming: π B1 A i1,...,b k A ik (R). Computations (e.g., string concatenation via ) to derive the value in new columns, e.g.: π SID,NAME FIRST LAST (STUDENTS). Such generalized π operators are also referred to as map operators (as in functional programming languages). Projection (6) 190 π A1,...,A k (R) corresponds to the SQL query: SELECT DISTINCT FROM A 1,...,A k R π B1 A 1,...,B k A k (R) is equivalent to the SQL query: SELECT DISTINCT FROM A 1 [AS] B 1,...,A k [AS] B k R

12 Selection vs. Projection 191 Selection vs. Projection Selection σ Projection π A 1 A 2 A 3 A 4 A 1 A 2 A 3 A 4 Filter some rows Maps all rows Combining Operations (1) 192 Since the result of any relational algebra operation is a relation again, this intermediate result may be the input of a subsequent RA operation. Example: retrieve the exercises solved by student with ID 102: π CAT,ENO (σ SID=102 (RESULTS)). We can think of the intermediate result to be stored in a named temporary relation (or as a macro definition): S102 σ SID=102 (RESULTS); π CAT,ENO (S102)

13 Combining Operations (2) 193 Composite RA expressions are typically depicted as operator trees: π CAT,ENO σ SID=102 + y RESULTS x 2 In these trees, computation proceeds bottom-up. The evaluation order of sibling branches is not pre-determined. Combining Operations (3) 194 SQL-92 permits the nesting of queries (the result of a SQL query may be used in a place of a relation name): Nested SQL Query SELECT DISTINCT CAT, ENO FROM (SELECT * FROM RESULTS WHERE SID = 102) AS S102 Note that this is not the typical style of SQL querying.

14 Combining Operations (4) 195 Instead, a single SQL query is equivalent to an RA operator tree containing σ, π, and (multiple) (see below): SELECT-FROM-WHERE Block SELECT DISTINCT CAT, ENO FROM RESULTS WHERE SID = 102 Really complex queries may be constructed step-by-step (using SQL s view mechanism), S102 may be used like a relation: SQL View Definition CREATE VIEW S102 AS SELECT * FROM RESULTS WHERE SID = 102 Relational Algebra 196 Overview 1. Introduction; Selection, Projection 2. Cartesian Product, Join 3. Set Operations 4. Outer Join 5. Formal Definitions, A Bit of Theory

15 Cartesian Product (1) 197 In general, queries need to combine information from several tables. In RA, such queries are formulated using, the Cartesian product. Cartesian Product The Cartesian product R S of two relations R, S is computed by concatenating each tuple t R with each tuple u S. ( denotes tuple concatenation.) Cartesian Product (2) 198 Cartesian Product A B C D = A B C D Since attribute names must be unique within a tuple, the Cartesian product may only be applied if R, S do not share any attribute names. (This is no real restriction because we have π.)

16 Cartesian Product (3) 199 If t = (A 1 : a 1,..., A n : a n ) and u = (B 1 : b 1,..., B m : b m ), then t u = (A 1 : a 1,..., A n : a n, B 1 : b 1,..., B m : b m ). Cartesian Product: Nested Loops create a new temporary relation T ; foreach t R do foreach u S do insert t u into T ; od od return T ; Cartesian Product and Renaming 200 R S may be computed by the equivalent SQL query (SQL does not impose the unique column name restriction, a column A of relation R may uniquely be identified by R.A): Cartesian Product in SQL SELECT * FROM R, S In RA, this is often formalized by means of of a renaming operator ϱ X (R). If sch(r) = (A 1 : D 1,..., A n : D n ), then ϱ X (R) π X.A1 A 1,...,X.A n A n (R).

17 Join (1) 201 The intermediate result generated by a Cartesian product may be quite large in general ( R = n, S = m R S = n m). Since the combination of Cartesian product and selection in queries is common, a special operator join has been introduced. Join The (theta-)join R θ S between relations R, S is defined as R θ S σ θ (R S). The join predicate θ may refer to attribute names of R and S. Join (2) 202 ϱ S (STUDENTS) S.SID=R.SID ϱ R (RESULTS) S.SID S.FIRST S.LAST S. R.SID R.CAT R.ENO R.POINTS 101 Ann Smith H Ann Smith H Ann Smith M Michael Jones (null) 102 H Michael Jones (null) 102 H Michael Jones (null) 102 M Richard Turner H Richard Turner M 1 7 Note: student Maria Brown does not appear in the join result.

18 Join (3) 203 R θ S can be evaluated by folding the procedures for σ, : Nested Loop Join create a new temporary relation T ; foreach t R do foreach u S do if θ(t u) then insert t u into T ; fi od od return T ; Join (4) 204 Join combines tuples from two relations and acts like a filter: tuples without join partner are removed. Note: if the join is used to follow a foreign key relationship, then no tuples are filtered: Join follows a foreign key relationship (dereference) RESULTS SID=S.SID π S.SID SID,FIRST,LAST, (STUDENTS) There are join variants which act like filters only: left and right semijoin (, ): R θ S π sch(r) (R θ S), or do not filter at all: outer-join (see below).

19 Natural Join 205 The natural join provides another useful abbreviation ( RA macro ). In the natural join R S, the join predicate θ is defined to be an conjunctive equality comparison of attributes sharing the same name in R, S. Natural join handles the necessary attribute renaming and projection. Natural Join Assume R(A, B, C) and S(B, C, D). Then: R S = π A,B,C,D (σ B=B C=C (R π B B,C C,D(S))) (Note: shared columns occur once in the result.) Joins in SQL (1) 206 In SQL, R θ S is normally written as Join in SQL ( classic and SQL-92) SELECT SELECT FROM R,S or FROM WHERE θ R JOIN S ON θ Note: this left query is exactly the SQL equivalent of σ θ (R S) we have seen before. SQL is a declarative language: it is the task of the SQL optimizer to infer that this query may be evaluated using a join instead of a Cartesian product.

20 Algebraic Laws (1) 207 The join satisfies the associativity condition (R S) T R (S T ). In join chains, parentheses are thus superfluous: R S T. Join is not commutative unless it is followed by a projection, i.e., a column reordering: π L (R S) π L (S R). Algebraic Laws (2) 208 A significant number of further algebraic laws hold, which are heavily utilized by the query optimizer. Example: selection push-down. If predicate ϕ refers to attributes in S only, then σ ϕ (R S) R σ ϕ (S). Selection push-down Why is selection push-down considered one of the most significant algebraic optimizations? (Such effficiency considerations are the subject of Datenbanken II. )

21 A Common Query Pattern (1) 209 The following operator tree structure is very common: π A1,...,A k σ ϕ θ1 θ2 R 1 θn 1 R n R n 1 R 2 1 Join all tables needed to answer the query, 2 select the relevant tuples, 3 project away all irrelevant columns. A Common Query Pattern (2) 210 The select-project-join query π A1,...,A k (σ ϕ (R 1 θ1 R 2 θ2 θn 1 R n )) has the obvious SQL equivalent SELECT DISTINCT FROM WHERE AND A 1,...,A k R 1,...,R n ϕ θ 1 AND AND θ n 1 It is a common source of errors to forget a join condition: think of the scenario R(A, B), S(B, C), T (C, D) when attributes A, D are relevant for the query output.

22 Relational Algebra Quiz (Level: Novice) 211 STUDENTS SID FIRST LAST 101 Ann Smith Michael Jones (null) 103 Richard Turner Maria Brown... EXERCISES CAT ENO TOPIC MAXPT H 1 Rel.Alg. 10 H 2 SQL 10 M 1 SQL 14 RESULTS SID CAT ENO POINTS 101 H H M H H M H M 1 7 Relational Algebra Quiz (Level: Novice) 212 Formulate equivalent queries in RA 1 Print all homework results for Ann Smith (print exercise number and points). 2 Who has got the maximum possible number of points (MAXPT) for a homework? Print full name and homework number. 3 (Who has got the maximum number of points for all homework exercises?)

23 Self Joins (1) 213 Sometimes it is necessary to refer to more than one tuple of the same relation at the same time. Example: Who got more points than the student with ID 101 for any of the exercises? Two answer this query, we need to compare two tuples t, u of the relation RESULTS: 1 tuple t corresponding to the student with ID 101, 2 tuple u, corresponding to the same exercise as the tuple t, in which u.points > t.points. Self Joins (2) 214 This requires a generalization of the select-project-join query pattern, in which two instances of the same relation are joined (the attributes in at least one instances must be renamed first): S := ϱ X (RESULTS) ϱ Y (RESULTS) X.CAT=Y.CAT X.ENO=Y.ENO π X.SID (σ X.POINTS>Y.POINTS Y.SID=101 )(S) Such joins are commonly referred to as self joins.

24 Relational Algebra 215 Overview 1. Introduction; Selection, Projection 2. Cartesian Product, Join 3. Set Operations 4. Outer Join 5. Formal Definitions, A Bit of Theory Set Operations (1) 216 Relations are sets (of tuples). The usual family of binary set operations can also be applied to relations. It is a requirement that both input relations have the same schema. Set Operations The set operations of relational algebra are R S, R S, and R S (union, intersection, difference). A minimal set of operations Which of these set operations is redundant (i.e., may be derived using an alternative RA expression, just like )?

25 Set Operations (2) 217 R R S R S S R S S R Set Operations (3) 218 R S may be implemented as follows: Union create a new temporary relation T ; foreach t R do insert t into T ; od foreach t S do insert t into T ; od remove duplicates in T ; return T ;

26 Set Operations (4) 219 R S may be implemented as follows: Difference create a new temporary relation T ; foreach t R do remove false; foreach u S do remove remove (t = u); od if remove then insert t into T ; fi od return T ; Union (1) 220 In RA queries, a typical application for is case analysis. Example: Grading MPOINTS := π SID,POINTS (σ CAT= M ENO=1 (RESULTS))) π SID,GRADE A (σ POINTS 12 (MPOINTS)) π SID,GRADE B (σ POINTS 10 POINTS<12 (MPOINTS)) π SID,GRADE C (σ POINTS 7 POINTS<10 (MPOINTS)) π SID,GRADE F (σ POINTS<7 (MPOINTS))

27 Union (2) 221 In SQL, is directly supported: keyword UNION. UNION may be placed between two SELECT-FROM-WHERE blocks: SQL s UNION SELECT SID, A AS GRADE FROM RESULTS WHERE CAT = M AND ENO = 1 AND POINTS = 12 UNION SELECT SID, B AS GRADE FROM RESULTS WHERE CAT = M AND ENO = 1 AND POINTS = 10 AND POINTS 12 UNION... Set Difference (1) 222 Note: the RA operators σ, π,,, are monotic by definition, e.g.: R S = σ ϕ (R) σ ϕ (S). Then it follows that every query Q that exclusively uses the above operators behaves monotonically: Let I 1 be a database state, and let I 2 = I 1 {t} (database state after insertion of tuple t). Then every tuple u contained in the answer to Q in state I 1 is also contained in the anser to Q in state I 2. Database insertion never invalidates a correct answer.

28 Set Difference (2) 223 If we pose non-monotonic queries, e.g., Which student has not solved any exercise? Who got the most points for Homework 1? Who has solved all exercises in the database? then it is obvious that σ, π,,, are not sufficient to formulate the query. Such queries require set difference ( ). A non-monotonic query Which student has not solved any exercise? (Print full name (FIRST, LAST). (Example database tables repeated on next slide.) Example Database (recap) 224 STUDENTS SID FIRST LAST 101 Ann Smith Michael Jones (null) 103 Richard Turner Maria Brown... EXERCISES CAT ENO TOPIC MAXPT H 1 Rel.Alg. 10 H 2 SQL 10 M 1 SQL 14 RESULTS SID CAT ENO POINTS 101 H H M H H M H M 1 7

29 Set Difference (3) 225 A correct solution? π FIRST,LAST (STUDENTS SID SID2 π SID2 SID (RESULTS)) A correct solution? π SID,FIRST,LAST (STUDENTS π SID (RESULTS)) Correct solution! Set Difference (4) 226 A typical RA query pattern involving set difference is the anti-join. Given R(A, B) and S(B, C), retrieve the tuples of R that do not have a (natural) join partner in S (Note: sch(r) sch(s) = {B}): R (π B (R) π B (S)). (The following is equivalent: R π sch(r) (R S).) There is no common symbol for this anti-join, but R S seems appropriate (complemented semi-join).

30 Set Operations and Complex Selections 227 Note that the availability of, (and ) renders complex selection predicates superfluous: Predicate Simplification Rules σ ϕ1 ϕ 2 (Q) = σ ϕ1 (Q) σ ϕ2 (Q) σ ϕ1 ϕ 2 (Q) = σ ϕ1 (Q) σ ϕ2 (Q) σ ϕ (Q) = Q σ ϕ (Q) RDBMS implement complex selection predicates anyway Why? Relational Algebra Quiz (Level: Intermediate) 228 The RA quiz below refers to the Homework DB. Schema: RESULTS (SID STUDENTS,(CAT, ENO) EXERCISES, POINTS) STUDENTS (SID,FIRST,LAST, ) EXERCISES (CAT,ENO,TOPIC,MAXPT) Formulate equivalent queries in RA 1 Who got the most points (of all students) for Homework 1? 2 Which students solved all exercises in the database?

31 Union vs. Join 229 Find RA expressions that translate between the two Two alternative representations of the homework, midterm exam, and final totals of the students are: RESULTS 1 STUDENT H M F Jim Ford Ann Smith RESULTS 2 STUDENT CAT PCT Jim Ford H 95 Jim Ford M 60 Jim Ford F 75 Ann Smith H 80 Ann Smith M 90 Ann Smith F 95 Summary 230 The five basic operations of relational algebra are: 1 σ ϕ 2 π L Selection Projection Cartesian Product Union Difference Derived (and thus redundant) operations: Theta-Join θ, Natural Join, Semi-Join, Renaming ϱ, and Intersection.

32 Relational Algebra 231 Overview 1. Introduction; Selection, Projection 2. Cartesian Product, Join 3. Set Operations 4. Outer Join 5. Formal Definitions, A Bit of Theory Outer Join (1) 232 Join ( ) eliminates tuples without partner: A B a 1 b 1 a 1 b 2 B C b 2 c 2 = b 3 c 3 A B C a 2 b 2 c 2 The left outer join preserves all tuples in its left argument, even if a tuple does not team up with a partner in the join: A B a 1 b 1 a 1 b 2 B C b 2 c 2 = b 3 c 3 A B C a 1 b 1 (null) a 2 b 2 c 2

33 Outer Join (2) 233 The right outer join preserves all tuples in its right argument: A B a 1 b 1 a 1 b 2 B C b 2 c 2 = b 3 c 3 A B C a 2 b 2 c 2 (null) b 3 c 3 The full outer join preserves all tuples in both arguments: A B a 1 b 1 a 1 b 2 B C b 2 c 2 = b 3 c 3 A B C a 1 b 1 (null) a 2 b 2 c 2 (null) b 3 c 3 Outer Join (3) 234 R θ S create a new temporary relation T ; foreach t R do haspartner false; foreach u S do if θ(t u) then insert t u into T ; haspartner true; fi od if haspartner then insert t (null,..., null) into T ; }{{} # attributes in S fi od return T ;

34 Outer Join (4) 235 Example: Prepare a full homework results report, including those students who did not hand in any solution at all: STUDENTS SID=SID π SID SID,ENO,POINTS(σ CAT= H (RESULTS)) SID FIRST LAST SID ENO POINTS 101 Ann Smith Ann Smith Michael Jones (null) Michael Jones (null) Richard Turner Maria Brown... (null) (null) (null) Outer Join (5) 236 Join vs. Outer Join Is there any difference between STUDENTS RESULTS and STUDENTS RESULTS? (Can you tell without looking at the table states?) Note: Outer join is a derived operation (like, ), i.e., it can simulated using the five basic relational algebra operations. Consider R(A, B) and S(B, C). Then R S (R S) ( (R π A,B (R S)) {(C:null)} ) SQL-92 provides {FULL, LEFT, RIGHT} OUTER JOIN.

35 Relational Algebra 237 Overview 1. Introduction; Selection, Projection 2. Cartesian Product, Join 3. Set Operations 4. Outer Join 5. Formal Definitions, A Bit of Theory Definitions: Syntax (1) 238 Let the following be given: A set D of data type names and for each D D a set val(d) of values. A set A of valid attribute names (identifiers). Relational Database Schema A relational database schema S consists of a finite set of relation names R, and for every R R a relation schema sch(r). (We ill ignore constraints here.)

36 Definitions: Syntax (2) 239 The set of syntactically correct RA expressions or queries is defined recursively, together with the resulting schema of each expression. Syntax of RA (Base Cases) 1 R (relation name) For every R R, R is an RA expression with schema sch(r). 2 {(A 1 :d 1,..., A n :d n )} (relation constant) A relation constant is an RA expression if A 1,..., A n A, d i val(d i ) for 1 i n with D 1,..., D n D. The schema of this expression is (A 1 :D 1,..., A n :D n ). Definitions: Syntax (3) 240 Let Q be an RA expr. with schema s = (A 1 :D 1,..., A n :D n ). Syntax of RA (Recursive Cases) 3 σ Ai =A j (Q) for i, j {1,..., n} is an RA expression with schema s. 4 σ Ai =d(q) for i {1,..., n} and d val(d i ) is an RA expression with schema s. 5 π B1 A i1,...,b m A im (Q) for i 1,..., i m {1,..., n} and B 1,..., B m A such that B j B k for j k is an RA expression with schema (B 1 :D i1,..., B m :D im ).

37 Definitions: Syntax (4) 241 Let Q 1, Q 2 be an RA expressions with the same schema s. Syntax of RA (Recursive Cases) 6 Q 1 Q 2 and 7 Q 1 Q 2 are RA expressions with schema s. Let Q 1, Q 2 be RA expressions with schemas (A 1 :D 1,..., A n :D n ) and (B 1 :E 1,..., B m :E m ), respectively. Syntax of RA (Recursive Cases) 8 Q 1 Q 2 is an RA expression with schema (A 1 :D 1,..., A n :D n, B 1 :E 1,..., B m :E m ) if {A 1,..., A n } {B 1,..., B m } =. Definitions: Semantics (1) 242 Database State A database state I (instance) defines a relation I(R) for every relation name R in the database schema S. The result of a query Q, i.e., an RA expression, in a database state I is a relation. This relation is denoted by I[Q] and defined recursively corresponding to the syntactic structure of Q.

38 Definition: Semantics (2) 243 I[Q] If Q is a relation name R, then I[Q] := I(R). If Q is a constant relation {(A 1 :d 1,..., A n :d n )}, then I[Q] := {(d 1,..., d n )}. If Q has the form σ Ai =A j (Q 1 ), then I[Q] := {(d 1,..., d n ) I[Q 1 ] d i = d j } If Q has the form σ Ai =d(q 1 ), then I[Q] := {(d 1,..., d n ) I[Q 1 ] d i = d} If Q has the form π B1 A i1,...,b m A im (Q 1 ), then I[Q] := {(d i1,..., d im ) (d 1,..., d n ) I[Q 1 ]} Definition: Semantics (3) 244 I[Q] (continued) If Q has the form Q 1 Q 2, then I[Q] := I[Q 1 ] I[Q 2 ] If Q has the form Q 1 Q 2, then I[Q] := I[Q 1 ] I[Q 2 ] If Q has the form Q 1 Q 2, then I[Q] := { (d 1,..., d n, e 1,..., e m ) (d 1,..., d n ) I[Q 1 ], (e 1,..., e m ) I[Q 2 ]}.

39 Monotonicity 245 Smaller Database State A database state I 1 is smaller than (or equal to) a database state I 2, written I 1 I 2, iff I 1 (R) I 2 (R) for all relation names R R of schema S. Theorem: RA \{ } is monotonic If an RA expression Q does not contain the (set difference) operator, then the following holds for all database states I 1, I 2 : I 1 I 2 = I 1 [Q] I 2 [Q]. Formulate proof by induction on syntactic structure of Q ( structural induction ). Equivalence (1) 246 Equivalence of RA Expressions Two RA expressions Q 1 and Q 2 are equivalent iff they have the same (result) schema and for all database states I, the following holds: I[Q 1 ] = I[Q 2 ]. Examples: σ ϕ1 (σ ϕ2 (Q)) = σ ϕ2 (σ ϕ1 (Q)) (Q 1 Q 2 ) Q 3 = Q 1 (Q 2 Q 3 ) If A is an attribute in the result schema of Q 1, then σ A=d (Q 1 Q 2 ) = (σ A=d (Q 1 )) Q 2. Theorem: The equivalence of (arbitrary) relational algebra expressions is undecidable.

40 Limitations of RA (1) 247 Let R be a relation name and assume sch(r) = (A:D, B:D), i.e., both columns share the same data type D. Let val(d) be infinite. The transitive closure of I(R) is the set of all (d, e) val(d) val(d) such that there are n N, n 1, and d 0, d 1,..., d n val(d) with d = d 0, e = d n and (d i 1, d i ) I(R) for i = 1,..., n. a d b c R from a b c to b c d Limitations of RA (2) 248 Theorem: There is no RA expression Q such that I[Q] is the transitive closure of I(R) for all database states I. In the directed graph example, one self-join (of R with itself) is needed, to follow the edges in the graph: π S.from,T.to (ϱ S (R) S.to=T.from ϱ T (R)) An n-fold self-join will find all paths in the graph of length n + 1. To compute the transitive closure for arbitrary graphs, i.e., for all database states I, is impossible in RA.

41 Limitations of RA (3) 249 This of course implies that relational algebra is not computationally complete. There are functions from database states to relations (query results), for which we could write a program 4, but we will not be able to find an equivalent RA expression to do the same. However, this would have been truly unexpected and actually unwanted, because want a guarantee that query evaluation always terminates. This is guaranteed for RA. Otherwise, we would have solved the halting problem. 4 Pick your favourite programming language. Limitations of RA (4) 250 All RA queries can be evaluated in time that is polynomically in the size of the database state. This implies that certain complex problems cannot be formulated in relational algebra. For example, if you find a way to formulate the Traveling Salesman problem in RA, you have solved the famous P = NP problem. (With a solution that nobody expects; contact me to collect your PhD.) As the transitive closure example shows, even not all problems of polynomial complexity can be formulated in classical RA.

42 Expressive Power (1) 251 Relational Completeness A query language L for the relational model is called strong relationally complete if, for every DB schema S and for every RA epxression Q 1 with respect to S there is a query Q 2 L such that for all database states I with respect to S the two queries produce the same results: I[Q 1 ] = I[Q 2 ]. Read as: It is possible to write an RA-to-L query compiler. Expressive Power (2) 252 SQL is strong relationally complete. If we can even write RA-to-L as well as L-to-RA compilers, both query languages are equivalent. SQL and RA are not equivalent. SQL contains concepts, e.g., the aggregate COUNT, which cannot be simulated in RA. Equivalent Query Languages 1 Relational algebra, 2 SQL without aggregations and with mandatory duplicate elimination, 3 Tuple relational calculus, 4 Datalog (a Prolog variant) without recursion.